Rolling Cylinder (Part 2)

Classical Mechanics Level 3

This is a follow-up to the linked problem

A solid cylinder of mass $20 \, \text{kg}$ and radius $0.12 \, \text{m}$ , rotating with an initial angular speed of $125 \, \text{rad/s}$ , is placed lightly (without any transitional push) onto a horizontal table at position $x = 0$ .

The coefficient of kinetic friction between the cylinder and the table is described by $\mu = 0.15 + x$ , where $x$ is the horizontal position of the cylinder's center of mass as it moves.

How long (in seconds) does it take the cylinder to begin rolling without slipping?

Note: The ambient gravitational acceleration is $9.8 \, \text{m/s}^2$ . Assume that the cylinder translates in the positive $x$ direction

The answer is 0.9777.

1 solution

Karan Chatrath
May 18, 2019

Nice problem! Here is a very brief solution.

Note that the forces acting on the ball are:

Its own weight acting downwards, acting at the COM of the cylinder.
The normal reaction force at the line of contact of the cylinder with the ground.
Friction, in the positive X direction, at the line of contact of the cylinder with the ground.

The normal reaction happens to be equal to the weight of the cylinder since there is no vertical acceleration.

The equations of motion along X and about the COM are found as such:

Force in the X - direction:

$M\ddot{x} = \mu Mg$

Net torque about COM:

$\frac{MR^2}{2}\ddot{\theta} = -\mu MgR$

These simplify to:

$\ddot{x} = \mu g$

$\ddot{\theta} = \frac{-2\mu g}{R}$

Having computed the equations of motion, the next step is to solve them. Analytical solutions can be found. With the knowledge of the initial conditions, the equations are however solved numerically and it was checked at which instant the pure rolling condition is satisfied. The pure rolling condition is:

$\dot{x} = R\dot{\theta}$

The required answer is 0.977 seconds .

Rolling Cylinder (Part 2)

The answer is 0.9777.

1 solution

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